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JEE Mathematics Solved Question Paper 2019

Multiple Choice Questions

If $△_{1} = |\begin{matrix} x & \sin (θ) & \cos (θ) \\ - \sin (θ) & - x & 1 \\ \cos (θ) & 1 & x \end{matrix}| and |\begin{matrix} x & \sin (2 θ) & \cos (2 θ) \\ - \sin (2 θ) & - x & 1 \\ \cos (2 θ) & 1 & x \end{matrix}|$ , $x \neq 0$ ; then for all $θ \in (0, \frac{π}{2})$ :

$△_{1} - △_{2} = x (\cos (2 θ) - \cos (4 θ))$
$△_{1} + △_{2} = - 2 x^{3}$
$△_{1} + △_{2} = - 2 (x^{3} + x - 1)$
$△_{1} - △_{2} = - 2 x^{3}$

If $f (x) = [\frac{\sin (p + 1) x + \sin (x)}{x}, x < 0 q, x = 0 \frac{\sqrt{x^{2} + x} - \sqrt{x}}{x^{\frac{3}{2}}}, x > 0]$ is continuous at x = 0, then the ordered pair (p, q) is equal to :

$(- \frac{1}{2}, \frac{3}{2})$
$(- \frac{3}{2}, - \frac{1}{2})$
$(\frac{5}{2}, \frac{1}{2})$
$(- \frac{3}{2}, \frac{1}{2})$

Let f(x) = e^x - x and g(x) = x² - x, $\forall x \in R$ . Then the set of all x $\in$ R, where the function h(x) = (fog)(x) is increasing, is :

$[- \frac{1}{2}, 0] \cup [1, \infty)$
$[- 1, - \frac{1}{2}] \cup [\frac{1}{2}, \infty)$
$[0, \infty)$
$[0, \frac{1}{2}] \cup [1, \infty)$

If the system of linear equations x + y + z = 5, x + 2y + 2z = 6, x + 3y + $λz = μ, (λ, μ \in R)$ has infinitely many solutions, then the value of $λ + μ$ is

Let f R $\to$ R : be differentiable at c $\in$ R  and f(c) = 0. If g(x) = $|f (x)|$ , then at x = c , g is

not differentiable
differentiable if f'(c) = 0
not differentiable if f'(c) = 0
differentiable if f'(c) $\neq$ 0

If y = y(x) is the solution of the differential equation $\frac{dy}{dx} = (\tan (x) - y) \sec^{2} (x), x \in (- \frac{π}{2}, \frac{π}{2})$ , such that y(0) = 0, then $y (- \frac{π}{4})$ is equal to

$2 + \frac{1}{e}$
e - 2
$\frac{1}{2} - e$
$\frac{1}{e} - 2$

7.
If the length of perpendicular from point $(β, 0, β)$ , $(where β \neq 0)$ to the line $\frac{x}{1} = \frac{y - 1}{0} = \frac{z + 1}{- 1} is \sqrt{\frac{3}{2}}$ , then the value of $β$ is

- 1

- 2

1

2

- 1

Let P(t, 1, - 1 - t) be the foot of perpendicular from $A (β, 0, β)$ on $\frac{x}{1} = \frac{y - 1}{0} = \frac{z + 1}{- 1}$

AP having d. r. s. < $t - β, 1, - 1 - t - β$ > should be perpendicular to line with d. r. s. < 1, 0, - 1

$\Rightarrow t - β + 1 + t + β = 0 \Rightarrow t = - \frac{1}{2} ⊥ Distance = \sqrt{{(t - β)}^{2} + 1^{2} + (- 1 - t + β)^{2}} = \sqrt{\frac{3}{2}} \Rightarrow 2 {(β + \frac{1}{2})}^{2} + 1 = \frac{3}{2} \Rightarrow β + \frac{1}{2} = \pm \frac{1}{2} \Rightarrow β = 0, - 1 Since β \neq 0, we get β = - 1$

Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is :

$\frac{1}{17}$
$\frac{1}{10}$
$\frac{1}{11}$
$\frac{1}{12}$

If $\int \frac{dx}{{(x^{2} - 2 x + 10)}^{2}} = A (\tan^{- 1} (\frac{x - 1}{3}) + \frac{f (x)}{x^{2} - 2 x + 10}) + C$ where C is a constant of integration, then